X+xy+3y=3;2x-xy-y=2 - решите систему уровнения

X(1+y)+3y=3
x=(3-3y)/(1+y)
$2 \times \frac{3 - 3y}{1 + y} + \frac{3 - 3y}{1 + y} \times y - y = 2$
$\frac{6 - 6y + 3y - 3 {y}^{2} - y - {y}^{2} }{1 + y} = 2$
$\frac{ - 4 {y}^{2} - 4y + 6 }{1 + y} = 2$
$\frac{ - 2(2 {y}^{2} + 2y - 3)}{1 + y} = 2$
$- 2(2 {y}^{2} + 2y - 3) = 2(1 + y)$
$- 2 {y}^{2} - 2y + 3 = 1 + y$
$2 {y}^{2} + 3y - 2 = 0$
D = 9-4*2*(-2) = 25
y1=(-3+5)/4 = 1/2
x1=(3-3*1/2)/(1+1/2) = 3/2 : 3/2 = 1

y2=(-3-5)/4=-2
x2= (3-3*(-2))/(1-2)= -9